There is mathematics behind the 1989 Tour de France ! The $1989$ Tour was won by Greg Lemond (USA, $1961$ - ), who beat  Laurent Fignon (France, $1960$ - $2010$) by $8''$. Yes, eight seconds! The closest tour in history.
Let me recall a few rules concerning the time measurement. At each stage, a measurement is made for every group of riders (for safety reasons, riders arriving together should not fight for winning seconds). Each time is a natural number of seconds. Therefore, each measurement involves a round-off and can be viewed as the integral part (closest integer) of a real (random) variable. For instance, if a rider arrives alone and $X$ is his real time, then his official time is the integer $n$ such that $X\in(n-1/2,n+1/2]$. At the end of the race, each rider has a total official time $T$, and his real time (which nobody knows) belongs to the interval $(T-m/2,T+m/2]$ with $m$ the number of stages ($21$ in $1989$).
When comparing two riders (say Greg and Laurent), we should only consider the stages where they arrived in different groups (they are not suppose to fight at the end of the irrelevant stages otherwise). This happened $11$ times to Greg and Laurent. Disregarding the irrelevant stages, we can say that the real time of Greg belongs to $(T_g-11/2,T_g+11/2]$ and that of Laurent  belongs to $(T_\ell-11/2,T_\ell+11/2]$. With $T_\ell=T_g+8$, we see that these intervals overlap! There is thus a possibility that Laurent rode faster than Greg but lost because of the round-offs. Of course, this has a very tiny probability, because the event needs that the round-offs be close to maximal at each stage and always in favor of Greg. Unlikely, but

What is the probability that Greg won just because of round-offs ?

Let us assume that the round-offs are independent random variables, uniformly distributed over $(-1/2,1/2]$.
By the way, I am not chauvinist, and I admire Greg.
 A: The sum of independent identically distributed random variables follows the Irwin-Hall distribution. For $n$ variables that are uniformly distributed on (0,1), the probability density function is
$$f_X(x)=\frac{1}{2\left(n-1\right)!}\sum_{k=0}^{n}\left(-1\right)^k{n \choose k}\left(x-k\right)^{n-1}\text{sgn}(x-k).$$
So for $n=11$ you will have to integrate an explicit piecewise polynomial function.
Edit: Tony is right, we need not quite the integral of the Irwin-Hall distribution. Rather, we need the difference distribution of two random variables that are distributed according to the above distribution (each of the RVs representing the total round-off error for each cyclist). So we need
$$\int_0^3\left(f_X(x)\int_{x+8}^{11}f_X(y)dy\right)dx.$$
This is the probability that the difference between the round off errors is between 8 and 11. It is obtained by integrating over all permissible combinations (one error, other error).
A: Caveat: this is not a complete answer, just a call for Matlab gurus to jump in here.  
Let $Y=Y_1+\dots  + Y_{11}$, where each $Y_i$ is the uniform difference distribution.  Note that the uniform difference distribution is a nice piecewise linear function with domain $[-1,1]$.  We simply want $P(Y \geq 8)$.  
